熊猫回路优化

感谢您添加数据和示例输出。首先,我很确定你不能将其矢量化,因为每个计算都依赖于前一个计算的输出。所以这是我所能做的最好的。

你的方法出现了 0.116999 在我的机器上几秒钟

这个进来了 0.0039999 秒

不是矢量化的,但是它得到了一个很好的速度提升,因为使用一个列表来处理这个问题并在最后将它添加回数据帧会更快。

def myfunc(pos_pre, signal):
    if signal == 0:  # if the signal is negative
        # if the new col value above -0.2 is > -1 then subtract 0.2 from that value
        pos = pos_pre - 0.02
        if pos < -1:  # if the row above - 0.2 < -1 set the value of current row to -1
            pos = -1

    elif signal == 1:
        # if the row above + 0.2 < 1 then add 0.2 to the value of the current row
        pos = pos_pre + 0.02
        if pos > 1:  # if the value above + 0.1 > 1 set the current row to 1
            pos = 1

    return pos


''' set first position value because you aren't technically calculating it correctly in your method since there is no 
position minus 1... IE: it will always be 0.02'''
new_pos = [0.02]

# skip index zero since there is no position 0 minus 1
for i in range(1, len(df)):
    new_pos.append(myfunc(pos_pre=new_pos[i-1], signal=df['signal'].iloc[i]))

df['position'] = new_pos

输出:

df.position
0    0.02
1    0.04
2    0.06
3    0.08
4    0.10
5    0.12
6    0.14
7    0.16
8    0.18

不要使用循环。熊猫擅长矢量化操作,例如 signal == 0 :

pos_shift = df['position'].shift() - 0.02
m1 = df['signal'] == 0
m2 = pos_shift < -1

df.loc[m1 & m2, 'position'] = -1
df['position'] = np.where(m1 & ~m2, pos_shift, df['position'])

你可以写一些类似的东西 signal == 1 .

是的。在寻找性能时,您应该始终在底层numpy阵列上操作:

signal = df['signal'].values
position = df['position'].values
for i in range(len(df)):
    if signal[i] == 0:
        if position[i-1]-0.02 < -1:
            position[i] = -1
        else:
            position[i] = position[i-1]-0.02
    elif signal[i] == 1:
        if position[i-1]+0.02 > 1:
            position[i] = 1
        else:
            position[i] = position[i-1]+0.02

你会惊讶于性能的提高,通常是10倍或更多。

很可能有更好的方法,但这个方法也应该奏效:

df['previous'] = df.signal.shift()

def get_signal_value(row):
    if row.signal == 0:
        compare = row.previous - 0.02
        if compare < -1:
            return -1
        else:
            return compare
    elif row.signal == 1: 
        compare = row.previous + 0.01
        if compare > 1:
            return 1
        else:
            return compare

df['new_signal'] = df.apply(lambda row: get_signal_value(row), axis=1)